Quadratic Equations are used to find maximums and minimums for rectangular regions. An example of this type of problem would occur when a person, with a specific amount of fencing, wants to find the largest rectangular area that can be fenced off.

In order to solve quadratic equations involving maximums and minimums for rectangular regions, it is necessary to

Since A represents a quadratic equation () in terms of l, we will re-write A in function form with the exponents in descending order:

The graph of will be a parabola and, since , the parabola will have a maximum point as its vertex. The y-coordinate of the vertex will represent our greatest area. To proceed, we need to find the value of the x-coordinate of the vertex (that is, the value of l in our equation).

Therefore the largest area that the farmer could enclose would be a square where each side has a length 250 yards.

yards

Examples

The owner of a ranch decides to enclose a rectangular region with 140 feet of fencing. To help the fencing cover more land, he plans to use one side of his barn as part of the enclosed region. What is the maximum area the rancher can enclose?

What is your answer?

A farmer wishes to enclose a rectangular region bordering a river using 600 feet of fencing. He wants to divide the region into two equal parts using some of the fence material. What is the maximum area that can be enclosed with the fencing?

What is your answer?

Examples

A rancher has 1200 feet of fencing to enclose two adjacent rectangular corrals of equal lengths and widths as shown in the figure below. What is the maximum area that can be enclosed in the fencing?

A local grocery store has plans to construct a rectangular parking lot on land that is bordered on one side by a highway. There are 1280 feet of fencing available to enclose the other three sides. [Let x represent the length of the two parallel sides of fencing.] Find the dimensions that will maximize the area of the parking lot.

For problems of this type you must correctly draw and label a figure to illustrate the given information. Then you must find equations to represent the perimeter and area. The equation for the area will be a quadratic which when graphed will be a parabola "opening down." To maximize the area, you must find the y-coordinate of the parabola's vertex. To do this we first used the vertex formula to find the x-coordinate and then substituted it back into the area formula.